Question: Which integral gives the length of the graph of $y=\sqrt x$ between $x=a$ and $x=b$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_a^b\sqrt{x+\sqrt x}~dx$ (Choice B) B $ \int_a^b\sqrt{x+\dfrac1{2\sqrt x}}~dx$ (Choice C) C $ \int_a^b\sqrt{1+\dfrac1{2\sqrt x}}~dx$ (Choice D) D $ \int_a^b\sqrt{1+\dfrac1{4x}}~dx$
Explanation: Recall that the formula for arc length of $~f(x)~$ over $~[a, b]~$ is $ L=\int_a^b\sqrt{1+\big[f\,^\prime(x)\big]^2}~dx\,$. First calculate $~f\,^\prime(x)\,$. $~f(x)={\sqrt x}\Rightarrow f\,^\prime(x)=\dfrac1{2\sqrt x}$ Next, use the formula above to write the integral expression that gives the arc length in question and simplify. $ L=\int_{a}^b\sqrt{1+\bigg[\dfrac1{2\sqrt x}\bigg]^2}~dx=\int_{a}^b\sqrt{1+\dfrac1{4x}}~dx$